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March 29, 2017

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Consider all 2 digit numbers N such that the last 2 digits of N^2 are N itself. What is the sum of all these 2 digit numbers?

  • Algebra - ,

    the answer is 76

  • Algebra - ,

    sorry the answer should be 101

    I do like this:
    first consider the possible last digit of the number, then you have 0,1,5,6

    Let N be ab

    For number end with 0,
    the last two digits of N will be 00
    no numbers can be formed as 00 is not a
    2 digit numbers

    For number end with 1,
    it must satisfy the conditions:
    last digit of 2a=a
    no numbers can be formed

    For number end with 5,
    it must satisfy the conditions:
    last digit of 10a+2=a
    the number 25 can be formed

    For number end with ,
    it must satisfy the conditions:
    last digit of 12a+3=a
    the number 76 can be formed

    Sum of N=76+25=101

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